integral of (9x^2+5)/(3x)

解

\int \frac{9 x ^{2} + 5}{3 x} d x

\frac{1}{3} (\frac{9 x ^{2}}{2} + 5 ln ∣ x ∣) + C

解答ステップ

\int \frac{9 x ^{2} + 5}{3 x} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{9 x ^{2} + 5}{x} d x

拡張

\frac{9 x ^{2} + 5}{x} : 9 x + \frac{5}{x}

= \frac{1}{3} \cdot \int 9 x + \frac{5}{x} d x

総和規則を適用する:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int 9 x d x + \int \frac{5}{x} d x)

\int 9 x d x = \frac{9 x ^{2}}{2}

\int \frac{5}{x} d x = 5 ln ∣ x ∣

= \frac{1}{3} (\frac{9 x ^{2}}{2} + 5 ln ∣ x ∣)

定数を解答に追加する

= \frac{1}{3} (\frac{9 x ^{2}}{2} + 5 ln ∣ x ∣) + C